Equations and graphs of motion

The $x$-$t$ graph of a moving object is a horizontal line at $x = 5\ \text{m}$ for $10\ \text{s}$. It is shown in the figure.

1. a) What is the state of the object during this time?
2. b) What is its velocity? Justify your answer using the gradient of the graph.

The $x$-$t$ graph of a pedestrian is a straight line passing through the points $(0, 2)$ and $(4, 10)$, with $x$ in metres and $t$ in seconds. It is shown in the figure.

1. a) Calculate the gradient of the line and state what physical quantity it represents.
2. b) Write the position equation for the pedestrian.

A classmate shows you two $x$-$t$ graphs: in the first, the position increases along a sloped straight line; in the second, the position increases along a curve that gets steeper and steeper. They are shown in the figure.

1. a) Which one corresponds to uniform motion and which to accelerated motion?
2. b) Explain what happens to the velocity in each case.

The $v$-$t$ graph of a cyclist is a horizontal line at $v = 6\ \text{m/s}$ from $t = 0$ to $t = 5\ \text{s}$. It is shown in the figure.

1. a) What type of motion does it describe?
2. b) Calculate the distance travelled using the area under the graph.

The $v$-$t$ graph of a car is a straight line joining the points $(0, 0)$ and $(4, 12)$, with $v$ in $\text{m/s}$ and $t$ in seconds. It is shown in the figure.

1. a) Calculate the gradient of the line and state what physical quantity it represents.
2. b) What is its value along with its unit?

The $v$-$t$ graph of a car is a straight line joining the points $(0, 20)$ and $(5, 0)$, with $v$ in $\text{m/s}$ and $t$ in seconds. It is shown in the figure.

1. a) Is the car accelerating or braking? Justify your answer.
2. b) Calculate its acceleration from the gradient.

The $v$-$t$ graph of a car has three sections: from $(0, 0)$ to $(10, 20)$, from $(10, 20)$ to $(30, 20)$, and from $(30, 20)$ to $(40, 0)$, with $v$ in $\text{m/s}$ and $t$ in seconds. It is shown in the figure.

1. a) Calculate the acceleration for each section.
2. b) Calculate the distance travelled in each section using the areas (triangles and rectangle) under the graph.
3. c) What is the total distance?

The $v$-$t$ graph of a metro train between two stations is: it accelerates from $0$ to $15\ \text{m/s}$ in $30\ \text{s}$, maintains $15\ \text{m/s}$ for $60\ \text{s}$ and brakes to a stop at $15\ \text{s}$. It is shown in the figure.

1. a) Calculate the acceleration for each section.
2. b) Calculate the distance of each section and the total distance between the stations.
3. c) Calculate the average speed for the journey, rounded to one decimal place.

The $a$-$t$ graph of a moving object is a horizontal line at $a = 2\ \text{m/s}^2$ from $t = 0$ to $t = 6\ \text{s}$, as shown in the figure.

If the object starts from rest, calculate its velocity after $6\ \text{s}$ using the area under the $a$-$t$ graph.

The $a$-$t$ graph of an object starting from rest is: $a = 3\ \text{m/s}^2$ from $t = 0$ to $t = 4\ \text{s}$, and $a = 0$ from $t = 4\ \text{s}$ to $t = 10\ \text{s}$. It is shown in the figure.

1. a) Calculate the velocity at $4\ \text{s}$ using the area under the $a$-$t$ graph.
2. b) What is its velocity at $10\ \text{s}$? Justify your answer.
3. c) Draw the corresponding $v$-$t$ graph in your notebook.

You can do it on paper and send a photo with your answer.

The $x$-$t$ graph of a walking trip has three straight sections: from $(0, 0)$ to $(10, 50)$, from $(10, 50)$ to $(15, 50)$, and from $(15, 50)$ to $(25, 0)$, with $x$ in metres and $t$ in seconds. It is shown in the figure.

1. a) Calculate the speed for each section.
2. b) Interpret each section: what are the walkers doing in each one?
3. c) How long does the entire trip last?

The $x$-$t$ graph of an object is a straight line connecting the points $(0, 50)$ and $(10, 0)$, with $x$ in metres and $t$ in seconds. It is shown in the figure.

1. a) Write its position equation.
2. b) Explain what the initial position and the sign of the gradient mean.
3. c) Calculate the position when $t = 14\ \text{s}$ and explain the physical meaning of it being negative.
4. d) At what time is it at $x = -50\ \text{m}$?

The $x$-$t$ graph of a walk is: a straight line from $(0, 0)$ to $(4, 40)$ and another from $(4, 40)$ to $(8, 0)$, with $x$ in metres and $t$ in seconds. It is shown in the figure.

1. a) Calculate the velocity for each section.
2. b) Calculate the total distance travelled and the total displacement.
3. c) Calculate the average velocity (displacement divided by time) and the average speed (distance divided by time).
4. d) Explain why one of them comes out as zero and the other does not.

Measuring the position of a laboratory trolley gives these data: at $t = 0\ \text{s}$, $x = 3\ \text{m}$; at $t = 1\ \text{s}$, $x = 5\ \text{m}$; at $t = 2\ \text{s}$, $x = 7\ \text{m}$; at $t = 3\ \text{s}$, $x = 9\ \text{m}$.

1. a) Plot the points on a $x$-$t$ graph in your notebook and check that they lie on a straight line.
2. b) Deduce the position equation for the trolley.
3. c) Predict its position when $t = 10\ \text{s}$.

You can do it on paper and send a photo with your answer.

The position of a moving object, in SI units, is $x = 10 - 3 \cdot t$.

1. a) Calculate its position at $t = 0\ \text{s}$, $t = 1\ \text{s}$ and $t = 2\ \text{s}$.
2. b) Describe the $x$-$t$ graph: is it sloping upwards or downwards? What does the negative gradient indicate?
3. c) Calculate the time at which it passes through the origin, rounding to one decimal place.

Measuring the position of a trolley going down an inclined plane gives: at $t = 0\ \text{s}$, $x = 0\ \text{m}$; at $t = 1\ \text{s}$, $x = 1\ \text{m}$; at $t = 2\ \text{s}$, $x = 4\ \text{m}$; at $t = 3\ \text{s}$, $x = 9\ \text{m}$.

1. a) Verify that the data satisfy the relation $x = t^2$ and, by comparing it with $x = \frac{1}{2} a \cdot t^2$, deduce the acceleration of the trolley.
2. b) Predict its position when $t = 5\ \text{s}$.
3. c) Calculate how far it moves in each second ($0$-$1\ \text{s}$, $1$-$2\ \text{s}$, $2$-$3\ \text{s}$) and explain, from these values, why the $x$-$t$ graph is a curve and not a straight line.